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Unfortunately, I don't know how to do draw commutative diagrams in TeX so I'll hope you're familiar with the statement of the lemma.

We want to show of course that $$0 \rightarrow \ker(f) \rightarrow \ker(g) \rightarrow \ker(h) \rightarrow \newcommand{\coker}{\operatorname{coker}}\coker(f) \rightarrow \coker(g) \rightarrow \coker(h) \rightarrow 0$$ is exact (where $f, g, h$ are as usual as in the statement of the lemma). The challenge of the snake's lemmma is to find the map $\ker(h) \rightarrow \coker(f)$ and show exactness at this step. But I'm afraid to say I don't see how even $0 \rightarrow \ker(f) \rightarrow \ker(g) \rightarrow \ker(h)$ and $\coker(f) \rightarrow \coker(g) \rightarrow \coker(h) \rightarrow 0$ is exact. As I stated, I would like to see the proof WITHOUT using elements, so without using Mitchell's embedding theorem, working in a general abelian category.

I know it arises 'naturally' from the fact that the sequences in the hypothesis of the lemma are exact, but I'm not sure when I actually try.

cqfd
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SES
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2 Answers2

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For this specific portion of the Snake Lemma, you can use the following fact :

Lemma : Let $$\require{AMScd}\begin{CD}A @>{k}>> B @>{f}>> C \\ @V{\alpha}VV (1) @VV{\beta}V @VV{\gamma}V \\ Ker(f') @>>{ker(f')}> B' @>>{f'}> C' \end{CD}$$ be a commutative diagram in a pointed category, with $\gamma$ a monomorphism. Then the square $(1)$ is a pullback if and only if $k=ker(f)$.

The short proof consists in noticing that because $\gamma$ is a monomorphism, $ker(f)=ker(\gamma f)=ker(f'\beta)$, and then apply the pullback lemma to the diagram $$\require{AMScd}\begin{CD}A @>{\alpha}>> Ker(f) @>>> 0 \\ @V{k}VV (1) @VV{ker(f')}V @VVV \\ B @>>{\beta}> B' @>>{f'}> C', \end{CD}$$ or alternatively prove directly that the universal properties for the pullback and the kernel coincide.

Once you have this lemma, you can look at the diagram $$\begin{CD} 0 @>>> Ker(f)@>{\tilde{k}}>> Ker(g) @>{\tilde{p}}>> Ker(h)\\ & & @V{ker(f)}VV (2) @VV{ker(g)}V @VV{ker(h)}V \\ 0 @>>> X @>{k}>> Y @>{p}>> Z \\ & @V{f}VV @VV{g}V @VV{h}V \\ 0 @>>> X' @>>{k'}> Y' @>>{p'}> Z'.\end{CD}$$ Since the lower row is exact, $k'$ is a monomorphism, and then the lemma implies that $(2)$ is a pullback. Since $ker(h)$ is also a monomorphism, the other impliciation now shows that $\tilde{k}$ is the kernel of $\tilde{p}$ and thus the top row is exact. The sequence with cokernels is also exact by duality.

Arnaud D.
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A complete proof without elements is given in the Stacks Project.

Fred Rohrer
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